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Calculus I
© 2007 Paul Dawkins
1
http://tutorial.math.lamar.edu/terms.aspx
Preface
Here are the solutions to the practice problems for my Calculus I notes. Some solutions will have
more or less detail than other solutions. The level of detail in each solution will depend up on
several issues.
If the section is a review section, this mostly applies to problems in the first
chapter, there will probably not be as much detail to the solutions given that the problems really
should be review.
As the difficulty level of the problems increases less detail will go into the
basics of the solution under the assumption that if you’ve reached the level of working the harder
problems then you will probably already understand the basics fairly well and won’t need all the
explanation.
This document was written with presentation on the web in mind.
On the web most solutions are
broken down into steps and many of the steps have hints.
Each hint on the web is given as a
popup however in this document they are listed prior to each step.
Also, on the web each step can
be viewed individually by clicking on links while in this document they are all showing.
Also,
there are liable to be some formatting parts in this document intended for help in generating the
web pages that haven’t been removed here. These issues may make the solutions a little difficult
to follow at times, but they should still be readable.
Limits
At
Infinity,
Part
II
1. For
()
3
82
xx
fx
+
=
e
evaluate each of the following limits.
(a)
lim
x
ﬁ¥
(b)
lim
x
ﬁ¥
(a)
lim
x
First notice that,
( )
3
lim
x
=¥
If you aren’t sure about this limit you should go back to the previous section and work some of
the examples there to make sure that you can do these kinds of limits.
Now, recalling
Example 1
from this section, we know that because the exponent goes to infinity
in the limit the answer is,
3
lim
x
e
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View Full Document Calculus I
© 2007 Paul Dawkins
2
http://tutorial.math.lamar.edu/terms.aspx
(b)
( )
lim
x
fx
ﬁ¥
First notice that,
( )
3
lim
82
x
xx
+
= ¥
If you aren’t sure about this limit you should go back to the previous section and work some of
the examples there to make sure that you can do these kinds of limits.
Now, recalling
Example 1
from this section, we know that because the exponent goes to negative
infinity in the limit the answer is,
3
li
m0
x
=
e
2. For
()
2
6
53
x
+
+
=
e
evaluate each of the following limits.
(a)
( )
lim
x
ﬁ¥
(b)
( )
lim
x
(a)
( )
lim
x
First notice that,
2
6
lim
x
x
+
+
If you aren’t sure about this limit you should go back to the previous section and work some of
the examples there to make sure that you can do these kinds of limits.
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This note was uploaded on 02/05/2011 for the course CALCULUS/C 202 taught by Professor Tadius during the Spring '11 term at Benedictine KS.
 Spring '11
 Tadius

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